# Using perturbation theory or small oscillation approximation in Harmonic oscillator

Let us assume, we are given the following potential,

$$V(x)=\frac{1}{2}ax^2-2x+\epsilon x^3$$

We need to find the energy levels of a particle bound in this potential

Let us think of the ground level for now. As we can see, the first term in the potential resembles a harmonic oscillator, so I can possibly use perturbation theory on the second and the third terms to come up with an approximate value of the ground level energy of this particle.

We'd get a ground state energy of the form,

$$E_0 = \frac{1}{2}\hbar \sqrt{\frac{a}{m}} + E_x^1 + E_x^3$$

The two terms, in the end, come from the *perturbations due to the linear and cubic terms.

We are usually given an arbitrary potential, and we approximate it into an oscillator potential about the local minima, using

\begin{align} \left.\frac{dV(x)}{dx}\right|_{x_{0}} &= 0 \\[5pt] \left.\frac{d^2V(x)}{dx^2}\right|_{x_{0}} &= K \\[5pt] \omega = \sqrt{\frac{K}{m}} \end{align}

This is how we obtain the value of angular frequency for a random potential, in small oscillations.

Now, the potential in our given problem can also be approximated like in small oscillations. We can obtain the minima at around $$x=\frac{\sqrt{a^2+24 \epsilon }-a}{6 \epsilon }\approx 2/a -12\epsilon/a$$, and by differentiating the potential twice, we get :

$$K=a+\frac{12\epsilon}{a}.$$

This equates to an angular frequency $$\omega = \sqrt{\frac{\frac{a^2+12\epsilon}{a}}{m}}$$

Using this frequency, we can also calculate the ground state energy level to be :

$$E_0 = \frac{1}{2}\hbar \sqrt{\frac{\frac{a^2+12\epsilon}{a}}{m}}$$

Thus using two different methods, perturbation theory, and small oscillations, I've obtained two approximations of the ground state energy level.

Which of the above methods is wrong, and why can I not use that ?

• What do you get for $E_x^1$ and $E_x^3$? Commented May 29, 2021 at 22:48
• @QuantumMechanic I've considered first order perturbation for this, and $E_x^1 = 0$ and $E_x^3 = 0$, because of odd symmetry. However, I've not yet learnt second order perturbation theory , but I know I'm supposed to get additional terms. I just want to know, why can I not use small oscillation approximation of potential, instead of perturbation theory. Commented May 29, 2021 at 23:05
• Why do you think one method must be wrong? You haven't evaluted one of them, they could yet give the same answer. Commented May 30, 2021 at 0:56
• There are several issues with your proposed solution. First is that the minimum is not near $x=1$ but rather at $x=\frac{\sqrt{a^2+24 \epsilon }-a}{6 \epsilon }\approx 2/a -12\epsilon/a$. The second derivative evaluated at that point yields \begin{align} k=\sqrt{a^2+24\epsilon} \approx a+\frac{12\epsilon}{a}\, . \end{align} Another difficulty is that $\epsilon$ is not dimensionless so you can't really use an expansion in powers of $\epsilon$ as this will results in terms that do not have dimensions of energy. Even with this I'm not sure why I can't get the two methods to agree. Commented May 30, 2021 at 2:05
• It is not clear to me how you can assume that the "linear" term is a perturbation of the quadratic term, as there is not "small" parameter multiplying it. I wonder if it might be better to rewrite your potential as $$V(x) = \frac{1}{2} a \left( x - \frac{2}{a} \right)^2 - \frac{2}{a} + \epsilon x^3,$$ and then treat the $\epsilon$ term as the perturbation. Commented May 30, 2021 at 6:37